a, Ta có : n² + 5n + 9 \(⋮\)n + 5

         = n (n + 5) + 9 \(⋮\)n + 5

    Vì n (n + 5) \(⋮\)n + 5 => 9 \(⋮\)n + 5

=> n + 5 \(\in\)Ư(9) = {\(\pm\)1 ; \(\pm\)3\(\pm\)9 }

Ta lập bảng :

Vậy ...................................................

Thấy đúng thì t.i.c.k đúng cho mik nhé !

\(M=\frac{-2x}{3}+3x\left(\frac{x}{6}-\frac{-2}{9}-\frac{7}{5}\right)-\frac{5x}{2}\left(\frac{x}{5}-\frac{4}{5}\right)\)

\(M=\frac{-2x}{3}+3x\left[\frac{x}{6}-\left(-\frac{2}{9}\right)-\frac{7}{5}\right]-\frac{5x}{4}\left(\frac{x}{5}-\frac{4}{5}\right)\)

\(M=\frac{-2x}{3}+3x\left(\frac{x}{6}+\frac{2}{9}-\frac{7}{5}\right)-\frac{5x}{2}\left(\frac{x}{5}-\frac{4}{5}\right)\)

\(M=-\frac{2x}{3}+3x\left(\frac{x}{6}-\frac{53}{45}\right)-\frac{5x}{2}.\frac{x-4}{5}\)

\(M=-\frac{2x}{3}+3x\left(\frac{x}{6}-\frac{53}{45}\right)-\frac{5x\left(x-4\right)}{10}\)

\(M=-\frac{2x}{3}+3x\left(\frac{x}{6}-\frac{53}{45}\right)-\frac{x\left(x-4\right)}{2}\)

\(M=-\frac{2x}{3}+\frac{x^2}{2}-\frac{53x}{15}-\frac{x\left(x-4\right)}{2}\)

\(M=\left(-\frac{2x}{3}-\frac{53x}{15}\right)+\frac{x^2}{2}-\frac{x\left(x-4\right)}{2}\)

\(M=-\frac{21x}{5}+\frac{x^2}{2}-\frac{x\left(x-4\right)}{2}\)

\(M=\frac{-2.21x+5x^2-5x\left(x-4\right)}{10}\)

\(M=\frac{-42x+5x^2-5x\left(x-4\right)}{10}\)

\(M=\frac{-x\left[42-5x+5\left(x-4\right)\right]}{10}\)

\(M=\frac{-x\left(42-5x+5x-20\right)}{10}\)

\(M=\frac{-x\left(42-20\right)}{10}\)

\(M=\frac{-x.22}{10}\)

\(M=-\frac{22x}{10}\)

\(M=-\frac{11x}{5}\)

\(\left|2x-\frac{1}{2}\right|>\left|-1,5\right|\)

\(\Rightarrow\left|2x-\frac{1}{2}\right|>1,5\)

\(\text{Vì }\left|2x-\frac{1}{2}\right|\ge0\)

\(\Rightarrow2x-\frac{1}{2}>1,5\)

\(\Rightarrow2x-\frac{1}{2}>\frac{15}{10}=\frac{3}{2}\)

\(\Rightarrow2x>2\)

\(\Rightarrow x>1\)

\(\frac{72-x}{3}=\frac{x-18}{5}\)

\(\Rightarrow\frac{\left(72-x\right).5}{15}=\frac{3\left(x-18\right)}{15}\)

\(\Rightarrow\left(72-x\right).5=3\left(x-18\right)\)

\(\Rightarrow360-5x=3x-54\)

\(\Rightarrow-3x-5x=-360-54\)

\(\Rightarrow-8x=-414\)

\(\Rightarrow x=51,75\)

+) Vì \(\frac{72-x}{3}=\frac{x-18}{5}\)

\(\Rightarrow5\left(72-x\right)=3\left(x-18\right)\)

     \(5.72-5x=3x-3.18\)

      \(360-5x=3x-54\)

       \(-5x-3x=-54-360\)

       \(-8x=-414\)

              \(x=-414:\left(-8\right)\)

              \(x=51,75\)

      Vậy x = 51,75

Ta có: 2x - 6 = 0 => x = 3

           x + 3 = 0 => x = -3

Lập bảng xét dấu:

+) Với x < -3

Ta có: 6 - 2x - x - 3 = 8

=> -3x + 3 = 8

=> -3x = 5

=> x = -5/3 (ko thỏa mãn) 

+) Với -3 ≤ x < 3

Ta có: 6 - 2x + x + 3 = 8

=> -x + 9 = 8

=> x = -1

=> x = 1 (thỏa mãn)

+) Với 3 ≤ x

Ta có: 2x - 6 + x + 3 = 8

=> 3x - 3 = 8

=> 3x = 11

=> x = 11/3 (thỏa mãn)

Vậy...

\(\frac{6}{x^2+2}+\frac{12}{x^2+8}=3-\frac{7}{x^2+3}\)

\(\Leftrightarrow6\left(x^2+8\right)\left(x^3+3\right)+12\left(x^2+2\right)\left(x^2+3\right)=3\left(x^2+2\right)\left(x^2+8\right)\left(x^2+3\right)-7\left(x^2+8\right)\left(x^2+2\right)\)

\(\Leftrightarrow18x^4+126x^2+216=3x^6+32x^4+62x^2+32\)

\(\Leftrightarrow18x^2+126x^2+216-3x^6-32x^4-68x^2-32=0\)

\(\Leftrightarrow-14x^4+58x^2+184-3x^6=0\)

\(\Leftrightarrow x=\pm2\)

\(\Rightarrow x=\pm2\)

a) \(4x^2\left(x-2\right)-x+2=0\)

\(\Leftrightarrow4x^2\left(x-2\right)-\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(4x^2-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\4x^2-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x=\pm\frac{1}{2}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=2\\x=\pm\frac{1}{2}\end{cases}}\)

g) \(x^2-7x+10=0\)

\(\Leftrightarrow\left(x-5\right)\left(x-2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-5=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=5\\x=2\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=5\\x=2\end{cases}}\)